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Consider the traditional AKM model where $$ Y_{it}=X_{it}\beta+\psi_{j(i,t)}+\epsilon_{it} $$ for $i=1,...,N$ (individual index), $t=1,...,T$ (time index), $j=1,...,J$ (firm index), and $j(i,t)$ is the firm employing worker $i$ at time $t$. For simplicity, I have ignored workers' fixed effects. $\psi_{j(i,t)}$ is called firm $j(i,t)$'s fixed effect.

By stacking observations over time, the equation above can be rewritten as $$ \underbrace{Y_i}_{T\times 1}=\underbrace{X_i}_{T\times K} \underbrace{\beta}_{K\times 1}+\underbrace{F_i}_{T\times J} \underbrace{\psi}_{J\times 1}+\underbrace{\epsilon_i}_{T\times 1} $$ where the $F(t,j)$ is $1$ if worker $i$ was employed at firm $j$ in period $t$.

Assume that:

[A1] We have a sample of observations $\{Y_i, X_i, F_i\}_{i=1}^N$ i.i.d., for $N$ large and $T$ small.

[A2] $E(\epsilon_i| X_i, F_i)=0$ for each $i=1,...,N$.

I would like to show how $\beta$ and $\psi$ are identified. I could not find any "understandable and organic" identification proof. Could you suggest one? I report here what I understood from reading/combining different sources.


My identification attempt (incomplete)

Step 1: By [A2], we have that

$$ \begin{pmatrix} E(X_i' Y_i)\\ E(F_i' Y_i)\\ \end{pmatrix}=\begin{pmatrix} E(X_i' X_i) & E(X_i' F_i)\\ E(F_i' X_i) & E(F_i' F_i)\\ \end{pmatrix} \begin{pmatrix} \beta\\ \psi \end{pmatrix} $$

Step 2: By [A1], the matrices $\begin{pmatrix} E(X_i' Y_i)\\ E(F_i' Y_i)\\ \end{pmatrix}$ and $\begin{pmatrix} E(X_i' X_i) & E(X_i' F_i)\\ E(F_i' X_i) & E(F_i' F_i)\\ \end{pmatrix} $ are consistently estimable from sample analogues. Hence, they can be treated as known in the identification exercise.

Step 3: By combining steps 2-3, it follows that if $A_{pop}\equiv \begin{pmatrix} E(X_i' X_i) & E(X_i' F_i)\\ E(F_i' X_i) & E(F_i' F_i)\\ \end{pmatrix} $ is invertible, then $\beta$ and $\psi$ are identified.

Step 4: Under which conditions is $A_{pop}$ invertible?

My understanding is that, instead of answering this question, Abowd, Kramarz, and Margolis provide necessary and sufficient conditions under which the sample analogue of the equation in Step 1 has a unique solution with respect to $(\beta,\psi)$. That is, they provide conditions under which $$ \begin{pmatrix} X'Y\\ F' Y\\ \end{pmatrix}=\begin{pmatrix} X'X & X'F\\ F' X & F'F\\ \end{pmatrix}\begin{pmatrix} \beta\\ \psi \end{pmatrix} $$ has a unique solution with respect to $(\beta,\psi)$ where $Y$, $X$, and $F$ stack $Y_i$, $X_i$ and $F_i$ over individuals. I suppose that, by law of large numbers, if the sample analogue has a unique solution, then also the population equation will have a unique solution.

Step 5: Consider separating the sample in $G$ groups satisfying the connectivity requirement discussed here.

Suppose, for simplicity, that $J=5$, $T=2$, $N=5$. Also suppose that worker 1 is employed at firm 1 in period 1 and at firm 2 in period 2; worker 2 is always employed at firm 1; worker 3 is employed at firm 3 in period 1 and at firm 2 in period 2; worker 4 is always employed at firm 3; worker 5 is employed at firm 5 in period 1 and at firm 4 in period 2. Hence, $G=2$. In group 1, there are firms 1,2,3. In group 2, there are firms 4,5. This is the example discussed in Figure 1 here.

In turn, the sample analogue of the equation in Step 1 when rearranging as discussed at p.5 of here is $$ \text{[Equation $(*)$]} \hspace{1cm}\begin{pmatrix} X'Y\\ F_1' Y\\ F_2' Y \end{pmatrix}=\underbrace{\begin{pmatrix} X'X & X'F_1 & X' F_2\\ F_1' X & F_1'F_1 & 0_{3\times 2}\\ F_2' X & 0_{2\times 3} & F_2' F_2 \end{pmatrix}}_{A_{sample}}\begin{pmatrix} \beta\\ \psi\\ \end{pmatrix} $$ where $$ F_1'F_1\equiv \begin{pmatrix} 3&0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3 \end{pmatrix}, F_2' F_2\equiv \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$$

From this point I'm lost. In particular,

(1) The authors tell us that Equation $(*)$ has a unique solution with respect to $(\beta,\psi)$ if and only if one among $\{\psi_1,\psi_2,\psi_3\}$ and one among $\{\psi_4,\psi_5\}$ are set equal to a known value (for example, $\psi_3=0$ and $\psi_4=0$). Could you help me to formally show this claim? In particular, I want to show both necessity and sufficiency.

If I look at the example above, I don't see why we need such normalisation.

I do understand that there cannot be an intercept term in $X$ in order to identify $\psi$. For simplicity, let us suppose that there is no $X$ at all.

Then, Equation $(*)$ reduces to the system $$ \begin{cases} Y_{11}+Y_{21}+Y_{22}=3\psi_1\\ Y_{12}+Y_{32}=2\psi_2\\ Y_{31}+Y_{41}+Y_{42}=3\psi_3\\ Y_{51}=\psi_4\\ Y_{52}=\psi_5\\ \end{cases} $$ that has a unique solution with respect to $\psi$ without the need of further normalising.

(2) Why, at p.10 of here, the authors go back to "expected values" rather than working with "sample means", given that we are concerned about uniqueness of solution of Equation $(*)$?

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    $\begingroup$ You are confusing yourself because you fail to include individual fixed effects. Consider wage equations for individual $i=5$: First time period $Ey_{i1} = \phi_4 + \mu_5$ and second time period $Ey_{i2} = \phi_5 + \mu_5$. Two linear equations with 3 unknowns NOT IDENTIFIED. Set $\phi_4=0$ then wage at first firm identifies individual fixed effect and the difference the firm effect $\phi_5$. Intuition: When you include individual fixed effect you need moovers between firms for identification. Seperated groups exists when there are no moovers to connect them. $\endgroup$ – Jesper Hybel Jan 26 at 17:39
  • $\begingroup$ Thank you. Your comment was very useful. I realised that, without worker fixed effects, avoiding a constant term allows us to identify the firm fixed effects. If we also have worker fixed effects, avoiding the constant is NOT SUFFICIENT anymore for identification. In the presence of BOTH worker and firm fixed effects, normalising one fixed effect within each group and avoiding the intercept allows us to identify worker and firm fixed effects. $\endgroup$ – user3285148 Jan 26 at 19:33
  • $\begingroup$ May I ask you a last clarification: suppose we have covariates and we also want to identify $\beta$ (without intercept). Are the normalisations described above still SUFFICIENT to identify the firm and worker fixed effects? I think the answer is "NO" because the equations will now also contain $\beta$. Are the normalisations described above still NECESSARY to identify the firm and worker fixed effects? I think the answer is "YES". What do you think? $\endgroup$ – user3285148 Jan 26 at 19:53
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Think about a regression where you have two factors. Normally when you have a constant in the regression you would put one level of each factor to 0 as a reference level. Leaving out the constant you have two factors and put one reference level to 0 for one of these.

The situation is similar here in the sense that you have firm and individual level fixed effects, each being factors with usually many levels and what is needed is a reference level within each connected firm-employee group. What is new is the connectedness.

In practice lets assume that you are studying employment for a given country made up geographically of two Islands. All workers on one Island only ever works for firms on that Island, but changing between these firms. Workers on the other Island only work for firms on that Island, again changing between these firms. Each Island defines a group of workers and firms with no worker in one group ever working for any firm in the other group.

Within each group you can compare the difference between working for one firm versus another as well as the efefct of one individual relative to another. But because no worker ever change from a firm on one Island to another you cannot compare the fixed effects across groups.

Definition (Connected) A firm-employee group is connected, when the group contains all the workers who ever worked for any of the firms in the group and all the firms at which any of the workers were ever employed.

The text you link to have the algorithm needed to find these groups in a given employer/employee matched data set.

In practice you have to find these groups and set one firm or employee fixed effect equal to 0 within each group. Doing that insures identification of firm and worker fixed effects as well as standard beta/parameters, by insuring full rank of the matrix defining normal equation is of full rank and the solution for the parameter vector is unique.

However, in R using lfe package there is an automatic check for this and it will define reference levels for you. I do not know about stata but I guess it is the same.

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  • $\begingroup$ Thanks. This is useful but what I'm looking for is a formal, analytical proof that can complete mine. The narrative you provide is important but insufficient to complete my proof. $\endgroup$ – user3285148 Jan 26 at 9:32
  • $\begingroup$ Fair enough. However, if that is the case it may be a good idea to say what about the proof - starting on page 10 of the note you link to - you need help in understanding. $\endgroup$ – Jesper Hybel Jan 26 at 11:56
  • $\begingroup$ Thanks. I have added details. Could you advise? $\endgroup$ – user3285148 Jan 26 at 12:25

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